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I am using the where function in numpy to look for a one letter string in an array of strings.
For example:
I will look for 'U' in ['B' 'U' 'A' 'M' 'R' 'O'] and get the index of 'U'.
letter = 'U'
row = ['B', 'U', 'A', 'M', 'R', 'O']
letter_found = np.where(row == letter)
However when I am looking for a letter that isn't present in the array of strings I get an empty tuple that looks like this:
(array([], dtype=int64),)
I need to be able to detect when it does not find the letter I am looking for in the array.
I've tried the following:
if not letter_found:
print 'not found'
But this does not work. How can I detect that the tuple returned from the where function in numpy is empty? Is it because one of my variables is possibly the wrong type? I am new at python and programming in general.
338
The preferred way to check if any list, dictionary, set, string or tuple is empty in Python is to simply use an if statement to check it. def is_empty(any_structure): if any_structure: print('Structure is not empty.
One of the ways we can easily check if a tuple is empty in Python is with the Python len() function. The length of a tuple which is empty is 0. Checking to see if a tuple is empty using the Python len() function is shown in the following Python code. empty_tuple = () if len(empty_tuple) == 0: print("Tuple is empty!")
You can check if the list is empty in python using the bool() function. bool() function returns boolean value of the specified object. The object will always return True , unless the object is empty, like [] , () , {} .
Empty lists are considered False in Python, hence the bool() function would return False if the list was passed as an argument. Other methods you can use to check if a list is empty are placing it inside an if statement, using the len() methods, or comparing it with an empty list.
The nomeclature:
if some_iterable:
#only if non-empty
only works when something is empty. In your case, the tuple isn't actually empty. The thing the tuple contains is empty. So you might want to do the following:
if any(map(len, my_tuple)):
#passes if any of the contained items are not empty
as len on an empty iterable will yield 0 and thus will be converted to False.
120
Your test is failing because letter_found is actually a tuple containing one element, so it's not empty. numpy.where returns a tuple of index values, one for each dimension in the array that you're testing. Typically when using this for searching in one-dimensional arrays, I use Python's tuple unpacking to avoid just this sort of situation:
letter = 'U'
row = ['B', 'U', 'A', 'M', 'R', 'O']
letter_found, = np.where(row == letter)
Note the comma after letter_found. This will unpack the result from numpy.where and assign letter_found to be the first element of that tuple.
Note also that letter_found will now refer to a numpy array, which cannot be used in a boolean context. You'll have to do something like:
if len(letter_found) == 0:
print('not found!')
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